Let $N\in \mathbb{N}$. I am looking for a non-zero scalar $\lambda$ and a nonzero vector
$$f=(f(0),f(1),\cdots,f(N-1)) \in \mathbb{C}^N$$ satisfying the following equations for $l=0,\cdots,N-1$:
$$\hat{f}(l)=\lambda e^{\frac{l\pi}{N}i}f(N-l).$$
$\hat{f}$ is just the Fourier transform of $f$ and it is assumed $f(N)=f(0)$.
A Gaussian might work. If you do $x[n] = e^{-jn^2/\sigma}$ with $\sigma = N/\pi$ you get $X(k) = \sqrt{N}x[k]$ assuming it's a periodic sequence anyway and you run your time & frequency indices from $-N/2$ to $N/2-1$
Since it's symmetric in both domains you automatically get $X[k] = \sqrt{N}x[N-k]$ as well.
I'm not quite sure what to do with $e^{j\pi\frac{k}{N}} = (-1)^k$
Time shifting by N/2 will modulate the frequency response by $(-1)^k$ but then you have to shift the frequency response as well which modulates the time domain signal. So it's almost there but not quite. Maybe the right combination of modulation and shifting in both domains will do the trick.
Okay, I am going to answer this question in two parts. The first part will be in the continuous-time and continuous-frequency domains (the regular Fourier Transform). I will use the "ordinary frequency" definition (that electrical engineers like to use) of the continuous-time Fourier Transform:
$$ \mathscr{F}\Big\{x(t)\Big\} \triangleq X(f) = \int\limits_{-\infty}^{\infty}x(t)\, e^{-j 2 \pi f t} \,\mathrm{d}t$$
$$ \mathscr{F}^{-1}\Big\{X(f)\Big\} \triangleq x(t) = \int\limits_{-\infty}^{\infty}X(f)\, e^{j 2 \pi f t} \,\mathrm{d}f$$
Note the symmetry between the forward and inverse Fourier transforms. They are, in a very important sense, exactly the same. This is because $-j$ and $+j$ have the same claim to squaring to be $-1$. If all textbooks and literature was changed so that every occurrence of $+j$ was replaced with $-j$ (and vise versa), every equation, every theorem, every derived fact would be just as true.
Now there is this Duality Theorem that says if:
$$ X(f) = \mathscr{F}\Big\{x(t)\Big\} $$
then
$$ x(-f) = \mathscr{F}\Big\{X(t)\Big\} $$
Note that all we're doing is swapping the roles of $f$ and $t$ and applying a ($-$) minus sign to one or the other (above it's $f$). That minus sign is necessary because while $-j$ and $+j$ are qualitatively the same, they are negatives of each other and they are not zero.
So, adjusting the notation a little, if we define $x_n(\cdot)$ like this:
$$\begin{align} x_1(f) &= \mathscr{F}\Big\{x_0(t)\Big\} \\ x_2(f) &= \mathscr{F}\Big\{x_1(t)\Big\} \\ x_3(f) &= \mathscr{F}\Big\{x_2(t)\Big\} \\ x_4(f) &= \mathscr{F}\Big\{x_3(t)\Big\} \\ \end{align} $$
Then you will see that $x_4(t) = x_0(t)$ for any $x_0(\cdot)$ that isn't pathologically (or "funkily") defined.
So, given any $x_0(\cdot)$, and the relationships above, if you define
$$ x(t) \triangleq x_0(t) + x_1(t) + x_2(t) + x_3(t) $$
then the Fourier Transform does not change $x(t)$.
$$ x(f) = \mathscr{F}\Big\{x(t)\Big\} $$
The function is exactly the same, you just changed the $t$ to an $f$.
Now the next part will be adjusting this to the Discrete Fourier Transform, because your question is about the DFT.
So now we're changing this to the Discrete Fourier Transform and we're gonna reach only half as far and I will use the "unitary" DFT scaling (which is not common) so that the symmetry between the discrete-time and discrete-frequency domain remains:
$$ \mathscr{F}\Big\{x[n]\Big\} \triangleq X[k] = \frac{1}{\sqrt{N}} \sum\limits_{n=0}^{N-1}x[n]\,e^{-j 2 \pi \frac{nk}{N}} $$
$$ \mathscr{F}^{-1}\Big\{X[k]\Big\} \triangleq x[n] = \frac{1}{\sqrt{N}} \sum\limits_{k=0}^{N-1}X[k]\,e^{j 2 \pi \frac{nk}{N}} $$
Note the symmetry between the forward and inverse discrete Fourier transforms. Now the mapping above in both cases is from one periodic discrete sequence of length $N$ in one domain to another periodic discrete sequence of the same length $N$ in the reciprocal domain.
$$ x[n+N] = x[n] \qquad \forall n \in \mathbb{Z} $$ $$ X[k+N] = X[k] \qquad \forall k \in \mathbb{Z} $$
The Duality Theorem says if:
$$ X[k] = \mathscr{F}\Big\{x[n]\Big\} $$
then
$$ x[-k] = \mathscr{F}\Big\{X[n]\Big\} $$
or (because of periodicity)
$$ x[N-k] = \mathscr{F}\Big\{X[n]\Big\} $$
Note that all we're doing is swapping the roles of $k$ and $n$ and applying a ($-$) minus sign to one or the other (above it's $k$).
If we define $x_m[\cdot]$ like this:
$$\begin{align} x_1[k] &= \mathscr{F}\Big\{x_0[n]\Big\} \\ x_2[k] &= \mathscr{F}\Big\{x_1[n]\Big\} \\ \end{align} $$
Then you will see that $x_2[n] = x_0[-n]$ for any $x_0[\cdot]$,
So, given any $x_0[\cdot]$, and the relationships above, if you define
$$ x[n] \triangleq x_0[n] + x_1[n] $$
then the DFT does not change $x[n]$ except for reversing $n$.
$$ x[k] = \mathscr{F}\Big\{x[-n]\Big\} $$
The function is exactly the same, you just changed the $-n$ to $k$.
or (because of periodicity)
$$ x[N-k] = \mathscr{F}\Big\{x[n]\Big\} $$
